The 2018 South African Mathematics Olympiad — Problem 5

The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. I have been writing about some of the problems from the senior paper from 2018. A list of all of the problems can be found here.

Today we will look at the fifth problem from the 2018 South African Mathematics Olympiad:

Determine all sequences $a_1, a_2, a_3, \ldots$ of nonnegative integers such that $a_1 < a_2 < a_3 < \ldots$, and $a_n$ divides $a_{n - 1} + n$ for all $n \geq 2$.

Since the sequence $a_1, a_2, \ldots$ is strictly increasing, we know that $a_n \geq n - 1$ for all positive integers $n$. (We could prove this rigorously by induction.) This means that $a_{n - 1} + n \leq (a_n - 1) + (a_n + 1) = 2a_n$ for all $n$, and so we know that $a_{n - 1} + n$ is equal to either $a_n$, or to $2a_n$ for all positive integers $n$. Perhaps we should try to figure out exactly when it is equal to $a_n$, and when it is equal to $2a_n$. If we knew, for example, that we always have that $a_{n - 1} + n = a_n$, then we have reduced the problem to solving this recurrence relation.…

The South African Mathematics Olympiad is an annual mathematics competition for high-school students in South Africa. The competition is organised by the South African Mathematics Foundation, and comprises three rounds which increase in difficulty. The final round of the 2019 South African Mathematics Olympiad will take place on Thursday, 25 July, and the top ten junior (Grade 8 and 9) and senior (Grades 10—12) competitors will be invited to a prize-giving evening taking place on 14 September 2019. At the same time, the problem selection committee will meet to start setting the 2020 papers.

According to the SAMF, nearly 100000 students participated in the 2017 edition of the competition. The numbers for the 2019 competition are likely to be similar. These students all write the first round of the competition, which learners write at their individual schools in March every year. The papers are marked at the school, and any student with more than 50% is invited to participate in the second round of the competition.…

Why did we choose that range for theta when doing trig substitutions?

Remember when we are doing a trig substitution, for instance for an integral with:

$\sqrt{a^2-x^2}$

We said that we should choose $x=a\sin\theta$, which seemed reasonable, but we also said that $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. Where did this last bit come from?

Well, we want a couple of things to hold true. The first is that any substitution that we make, we have to be able to undo. That is, we will substitute $x$ for a function of $\theta$ but in the end we need to convert back to $x$ and so to do that we have to be able to write the inverse function of, in this case $x=a\sin\theta$. The $\sin$ function is itself not invertible because it’s not one to one, so we have to choose a range over which it is one to one. We could choose $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$ or we could choose $\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}$ (amongst an infinite set of possibilities). That would also be invertible. However, remember that we are going to end up with a term of the form:

$\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}$

So if we want this to simplify, we had better choose our range of $\theta$ such that $\cos\theta$ is positive, so that we can write $\sqrt{\cos^2\theta}=\cos\theta$.…

Integrals with sec and tan when the power of tan is odd

We went through an example in class today which was

$\int tan^6\theta \sec^4\theta d\theta$

In this case we took out two powers of sec and then converted all the other $\sec$ into $latex\ tan$, which left a function of tan times $sec^2\theta d\theta$. We wanted to do this because the derivative of $\tan$ is $\sec^2$ and so we can do a simple substitution. If we have an odd power of $\tan$, we can employ a different trick. Let’s look at:

$I=\int \tan^5\theta\sec^7\theta d\theta$.

Here, sec is an odd power and so we can’t employ the same trick as before. Now we want to convert everything to a function of $\sec$ and have only a factor which is the derivative of $\sec$ left over. The derivative of $\sec$ is $\sec\tan$, so let’s try and take this out:

$I=\int \tan^5\theta\sec^7\theta d\theta=\int \tan^4\theta\sec^6\theta (\sec\theta\tan\theta)d\theta$.

Now convert the $\tan$ into $\sec$ by $\tan^2\theta=\sec^2\theta-1$:

$I=\int (\sec^2\theta-1)^2\sec^6\theta (\sec\theta\tan\theta)d\theta=\int (\sec^{10}\theta-2\sec^8\theta+\sec^6\theta) (\sec\theta\tan\theta)d\theta$

where here we have just expanded out the bracket and multiplied everything out.…

Fundamental theorem of calculus example

We did an example today in class which I wanted to go through again here. The question was to calculate

$\frac{d}{dx}\int_a^{x^4}\sec t dt$

We spot the pattern immediately that it’s an FTC part 1 type question, but it’s not quite there yet. In the FTC part 1, the upper limit of the integral is just $x$, and not $x^4$. A question that we would be able to answer is:

$\frac{d}{dx}\int_a^{x}\sec t dt$

This would just be $\sec x$. Or, of course, we can show that in exactly the same way:

$\frac{d}{du}\int_a^{u}\sec t dt=\sec u$

That’s just changing the names of the variables, which is fine, right? But that’s not quite the question. So, how can we convert from $x^4$ to $u$? Well, how about a substitution? How about letting $x^4=u$ and seeing what happens. This is actually just a chain rule. It’s like if I asked you to calculate:

$\frac{d}{dx} g(x^4)$.

You would just say: Let $x^4=u$ and then we have:

$\frac{d}{dx} g(x^4)=\frac{du}{dx}\frac{d}{du}g(u)=4x^3 g'(u)$.…

Using integration to calculate the volume of a solid with a known cross-sectional area.

Hi there again, I have not written a post in while, here goes my second post.

I would like us to discuss one of the important applications of integration. We have seen how integration can be used to solve the area problem, in this post we are going to see how we can use a similar idea to solve the volume problem. I suggest that we start by looking at the solids whose volume we know very well. You should be able to calculate the volumes of the cylinders below (yes,  they are all cylinders.)

Cylinders are nice, we only need to multiply the cross-sectional area by the height/length to find the volume. This is because they have two identical flat ends and the same cross-section from one end to the other. Unfortunately, not all the solid figures that we come across everyday are cylinders. The figures below are not cylinders.…

Introduction to trigonometric substitution

I have decided to start writing some posts here, and this is my first post. I would like to introduce trig substitution by presenting an example that you have seen before. Trig substitution is one of the techniques of integration, it’s like u substitution, except that you use a trig function only.

Let’s get into the example already!

$\int_{-1}^{1} \sqrt{1-x^2} dx$

If you equate the integrand to y (and get $x^2+y^2=1$, $y\geq 0$), you should be able to see that this is the area of the upper half of a unit circle. The answer to this definite integral is therefore the area of the upper half of the unit circle (yes, the definite integral of f(x) from a to b gives you the net area between f(x) and the x-axis from x=a to x=b), is $\frac{\pi}{2}$.

We relied on the geometrical interpretation of the integral to solve the definite integral, but can we also show this algebraically?…

MAM1000W 2017 semester 2, lecture 1 (part ii)

The distance problem

If I want to know how far I walked during an hour, I can ask how far I walked in the first five minutes, and how far I walked in the second five minutes, and how far I walked in the third five minutes, etc. and add them all together. ie. I could write:

$d=d_1+d_2+d_3+d_4+...d_{12}$

Where $d_i$ is the distance walked in the $i^{th}$ five minutes. To calculate a distance, we need to know how fast we are going, and for how long. In fact:

$distance=velocity \times time$

where you can think of velocity as the same thing as speed (though there are subtle differences which you will find out about later). This formula works if the velocity is constant, but what if it is changing. Well, if we have a graph of velocity against time, then we can think about splitting the graph into intervals (like the five minute intervals above), and approximating that during a small interval of time, the velocity is roughly constant.…

MAM1000W 2017 semester 2, lecture 1 (part i)

I wanted to put up a little summary of some of the most important things to remember from the end of last semester. There was a sudden input of new concepts, so let’s put some of them down here to get a clear reminder of what we need to know. A few things in this post:

• The antiderivative
• Sigma notation
• Areas under curves

Antiderivatives

An antiderivative of a function $f$ on an open interval $I$ is a function $F$ such that:

$F'(x)=f(x)$ for every $x\in I$

Note that we say an antiderivative, not the antiderivative. There can be many functions whose derivatives give the same thing. While we know that:

$\frac{d}{dx}\sin x=\cos x$

and therefore  $\sin x$ is an antiderivative of $\cos x$, we can also say that:

$\frac{d}{dx}(\sin x+3)=\cos x$

So $\sin x+3$ is also an antiderivative of $\cos x$. In fact for any constant $c$ it is true that $\sin x+c$ is an antiderivative of $\cos x$. We will come up with some clever notation for the antiderivative soon.…

Unsolved!: The History and Mystery of the World’s Greatest Ciphers from Ancient Egypt to Online Secret Societies by Craig P. Bauer – A review

This book was sent to me by the publisher as a review copy.

This is a book of some impressive magnitude, both in terms of the time span that it covers (being millennia), as well as the ways in which it discusses the context and content of the ciphers, most of which, as the title suggests, are unsolved. The book starts with perhaps the most mysterious of all unbroken ciphers: The Voynich Manuscript (the entirety of which can be found here). This story in itself is perhaps the most fascinating in the history of all encrypted documents, and that we still don’t know if it truly contains anything of interest, or is just a cleverly constructed (though several hundred year old) hoax makes it all the more intriguing.

The writing rather effortlessly weaves between the potential origin stories, the history of the ownership of the manuscript and the attempts to decode it.…